Give $a,b,c>0$. Prove that: $$\dfrac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}.$$
My direction: (we have the equation if and only if $a=b=c$)
$a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt[3]{abc}$
$b^{n+1}+b^na+b^nc \ge 3b^n\sqrt[3]{abc}$
$c^{n+1}+c^na+c^nb \ge 3c^n\sqrt[3]{abc}$
But from these things, i can't prove the problem.
On
Observe that by AM-GM inequality for $3$ positive reals: $\dfrac{a+b+c}{3} \ge \sqrt[3]{abc}$. Thus you need to show: $\dfrac{a^{n+1}+b^{n+1} + c^{n+1}}{a^n+b^n+c^n} \ge \dfrac{a+b+c}{3}$, which is the same as: $2(a^{n+1}+b^{n+1}+c^{n+1}) \ge a^n(b+c)+b^n(c+a)+c^n(a+b)$. But this is quite clear because you can use the inequality: $(x-y)(x^n-y^n) \ge 0 \implies x^{n+1}+y^{n+1} \ge x^ny+xy^n$ three times for the pairs $(a,b), (b,c), (c,a)$ and add up.
On
WLOG let $a \geq b \geq c$ , then $\sqrt[3]a \geq \sqrt[3]b \geq \sqrt[3]c$. Let $x = \sqrt[3]a, y = \sqrt[3]b , z =\sqrt[3]c$ , then $x \geq y\geq z > 0$ and the inequality is equivalent to : $$ x^{3n+3} + y^{3n+3} + z^{3n+3} \geq x^{3n+1}yz + y^{3n+1}xz + z^{3n+1}xy $$
by Muirhead's inequality, since the sequence $(3n+3,0,0)$ majorizes $(3n+1,1,1)$, we know that $$ \sum_{\mathrm{sym}} x^{3n+3}y^0z^0 \geq \sum_{\mathrm{sym}} x^{3n+1}y^1z^1 $$
which upon division by $2$ gives the result we require. Note that $n$ natural is required for the inequality to hold.
On
Another way.
We need to prove that: $$\sum_{cyc}a^{n+1}\geq\sum_{cyc}a^{n+\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}},$$ which is true by Muirhead because $$(n+1,0,0)\succ\left(n+\frac{1}{3},\frac{1}{3},\frac{1}{3}\right).$$
On
By Chebyshev's sum inequality (https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality), we have $$a^n\cdot a + b^n\cdot b + c^n \cdot c \ge \frac{1}{3}(a^n + b^n + c^n)(a+b+c).$$ Thus, $$\frac{a^{n+1}+b^{n+1} + c^{n+1}}{a^n + b^n + c^n} \ge \frac{a+b+c}{3} \ge \sqrt[3]{abc}.$$ We are done.
On
Let $A_p:=\left(\frac{1}{N}\sum_{i=1}^Na_i^p\right)^{1/p}$ be the $p$th mean of $(a_i)$.
By the extended AM-GM inequality, $GM\le A_n\le A_{n+1}$. Hence $$GM\times A_n^n\le A_{n+1}\times A_{n+1}^n=A_{n+1}^{n+1}$$ or $$\sqrt[3]{abc}\times\frac{a^n+b^n+c^n}{3}\le\frac{a^{n+1}+b^{n+1}+c^{n+1}}{3}$$
Because for natural $n$ by AM-GM we obtain: $$\begin{aligned} \sum_{\text{cyc}}a^{n+1}&=\frac{1}{3(n+1)}\sum_{\text{cyc}}\left((3n+1)a^{n+1}+b^{n+1}+c^{n+1}\right) \\ &\geq \sum_{\text{cyc}}\sqrt[3n+3]{a^{(3n+1)(n+1)}b^{n+1}c^{b+1}} \\ &=\sum_{\text{cyc}}a^{n+\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}} \\ &=\sqrt[3]{abc}\sum_{\text{cyc}}a^n \end{aligned}$$